FACTORING AND SOLVING QUADRATIC LESSON EQUATIONS 11... · Solving quadratic equations by completing the square We can also solve equations of the form ax2 + bx + c = 0 by completing

Page 72Page 72

FACTORING AND SOLVING QUADRATIC EQUATIONS

Solving the equation ax2 + bx + c = 0

Example 1

Solve the equation x² + 8x +15 = 0.

The zero principle: When we solve quadratic equations, it is essential to get the one side of the equation equal to zero before we factorise the quadratic part of the equation. The zero principle is very important, since when the quadratic is factorised, it will form a product of the two factors that will be equal to zero. So now we see that when factoring we obtain:

x² + 8x +15 = 0

\ (x + 3)(x + 5) = 0

Now to complete the problem, we simply apply a very basic principle that applies to a product that is equal to zero. If we have two factors A and B such that A·B = 0, then only one of the two has to be zero for the product to be zero. That would mean that A = 0 or B = 0.

So now:

(x + 3)(x + 5) = 0

\ x = −3 or x = −5

Note that if this product A·B = 6, then we cannot be certain of the values of each of the factors A and B since 6 = 2 × 3 or 6 = 1 × 6 or 6 = 1 _ 2 × 12, etc. So it is important that we ensure the one side is zero.

Example 2

Fully factorise 3x² + 33x + 84 and hence solve 3x2 + 33x + 84 = 0.

To factorise:

3x² + 33x + 84 = 3(x² + 11x + 28)

= 3(x + 4)(x + 7)

When solving 3x² + 33x + 84 = 0 the zero principle applies. So:

3(x + 4)(x + 7) = 0

\ (x + 4)(x + 7) = 0 (We can divide both sides by 3)

\ x = −4 or x = −7

6LESSON6LESSON

ExampleExample

ExampleExample

Page 73Lesson 1 | Algebra

Page 12009Page 73

Activity 1

Solve for x in each of the following:

1. x² − 7 x +10 = 0 2. x² + 4x = 21

3. 12x² − 6 = −17 x −12 4. x² = 6 − 5x

5. 6x2 + 11x + 3 = 0 6. 8x2 – 14x + 3 = 0

7. 21x2 = 41x – 10 8. 3x2 + x = 14

Solving quadratic equations using the quadratic formula

If we cannot find the factors for a quadratic equation of the form ax2 + bx + c = 0, then it is quite possible that the roots are irrational.

So if ax2 + bx + c = 0 , then

x = − b ± √ __ b2 − 4ac __ 2a

We prove this by colmpeting the square:

ax2 + bx + c = 0 ax2 + bx + c = 0

( . _ . a): x2 + b _ a x + c _ a = 0 (x 4a): 4a2x2 + 4abx + 4ac = 0

Complete the square: Complete the square:

x2 + b _ a x + ( b _ 2a )2 = ( b _ 2a )2 − c _ a 4a2x2 + 4abx2 + b2 = b2 - 4ac

\(x + b _ 2a )2 = b2 _ 4a2 − c _ a \ (2ax + b)2 = b2 − 4ac

= b2 − 4ac _ 4a2 \ 2ax + b = ± √

_ b2 − 4ac

\ x+ b _ 2a = ± √ _ b

2 − 4ac _ 4a2 \ 2ax = −b ± √ _ b2 − 4ac

\ x = − b _ 2a ± √ _ b

2 − 4ac _ 2a x = −b ± √ _ b2 − 4ac __ 2a

x = −b ± √ __ b2 - 4ac __ 2a

ActivityActivity

Page 74Page 74Page 74Lesson 1 | Algebra

Page 1Page 74

You may be required to give the solutions for x correct to a specific number of decimal places, or alternatively in simplest surd form.

Although you can use the formula as an alternative to factorising an equation and then solving, make sure you can factorise quadratic expressions, as the next step will be factoring cubic polynomials. The formula only works on quadratic equations, and not on cubics. (More about this in Grade 12).

Example 1

Solve for x in 2x2 + x − 2 = 0

Remember the standard form

ax2 + bx + c = 0

So: a = 2; b = 1; c = −2

\ x = −b ± √ __ b2 − 4ac __ 2a

= −1 ± √ __ 1 − 4(−4) __ 4

= −1 ± √ _ 17 __ 4

So in simplest surd form:

\ x = − 1 + √ _ 17 __ 4 or − 1 − √

_ 17 __ 4

Correct to two decimal places:

x = 0,78 or x = −1,28

Example 2

Solve for x in (x + 3)(2 − x) = 7x.

We need to write the equation in standard form:

(x + 3)(2 − x) = 7x

\ 2x − x2 + 6 − 3x − 7x = 0

\ −x2 − 8x + 6 = 0

\ x2 + 8x − 6 = 0

Now a = 1 b = 8 c = −6

\ x = −b ± √ __ b2 − 4ac __ 2a

= −8 ± √ __ 64 − 4(−6) __ 2

= −8 ± √ _ 88 __ 2

= −8 ± 2 √ _ 22 __ 2

= 2(−4 ± √

_ 22 ) __ 2

= − 4 ± √ _ 22

So in simplest surd form:

\ x = − 4 − √ _ 22 or x = − 4 + √

_ 22

Correct to two decimal places:

x = −7,32 or x = −0,69

Activity 2

Solve for x in each of the following by using the formula. Leave your answers in simplest surd form and rounded to two decimal places where applicable:

1. 2x − 5 = 7 _ x 2. 1 _ 2 x2 − 3x + 0, 7 = 0

ExampleExample

Using the casio Fx 82 -

ES Plus: If you enter this

line on your calculator,

the answer that you get

will be in simplest surd

form.

Using the casio Fx 82 -

ES Plus: If you enter this

line on your calculator,

the answer that you get

will be in simplest surd

form.

ActivityActivity

Lesson 1 | Algebra

Page 1Page 75Lesson 1 | Algebra

Page 12009Page 75

3. 4x2 − 20x + 9 = 0 4. 2x2 + 4x + 3 = 0

Solving quadratic equations by completing the square

We can also solve equations of the form ax2 + bx + c = 0 by completing the square, which means rewriting the equation in the form a(x − p)2 + q = 0 . Do you notice that the x is contained in a bracket which is under a square. So it will be easier to solve as we can merely take the square root both sides.

Equations with squares

Example 1

Solve for x in 4(x − 3)2 = 16

We divide to isolate the bracket:

(x − 3)2 = 16 _ 4

\ (x − 3)2 = 4

Now square root both sides and make sure not to forget the “ ± “ sign:

\ x − 3 = ±2

And solve:

\ x = 3 ± 2

\ x = 5 or x = 1.

Example 2

Solve for x in 4(x − 3)2 = 5

We divide to isolate the bracket:

(x – 3)2 = 5 _ 4

Now square root both sides and make sure not to forget the “ ± “ sign:

\ x – 3 = ± √ _ 5 _ 2

And solve:

\ x = 3 ± √ _ 5 _ 2 in surd form

or rounded to two decimal places:

x = 1,88 or x = 4,12

ExampleExample

ExampleExample


Page 1Page 76

Example 3

Solve for x in 4(x − 3)2 = −5

Here it is clear to see that the left hand side of the equation which is 4(x − 3)2 is always positive. This happens because anything squared is positive. So (x − 3)2 ≥ 0. If we multiply this by a positive number, it remains positive. So 4(x − 3)2 ≥ 0. So the right hand side of the equation should be positive, which it is not. So this equation has no real solutions. We say that it has non-real roots

since solving it would mean, therefore, that x − 3 = ± √ _ −5 _ 2 and we know that √

_ −5

is a non-real number.

It is clear to see that if the x is contained in a bracket, we can say a lot about the value of the expression, and we can also solve it more easily. So let us look at a few methods that help us complete the square.

Completing the square to solve the equation ax2 + bx + c = 0

Example 1

In the following example we will solve the equation x2 − 2x − 24 = 0 by completing the square.

We will do two different methods, and you can choose which one to follow:

Method 1 Method 2Solve for x in x2 − 2x − 24 = 0 Solve for x in x2 − 2x − 24 = 0To complete the square we must follow the steps below:1. Ensure that a in ax2 + bx + c = 0 is equal to one2. Move the constant term(s) to the opposite side of the equation.3. To complete the square, we

add ( b _ 2 ) 2 to both sides of the

equation.4. Factorise and isolate the completed square.5. Take the square root of both sides.6. Isolate x and solve.

To complete the square we must follow the steps below:1. Move the constant term to the opposite side of the equation.2. Take the value of a and multiply it by 4 so you create 4a.3. Multiply each term by 4a.4. Calculate b2 and add it to both sides (note: the b value that you use is determined from the original equation).5. Factorise the expression (in the form of a perfect square).6. Take the square root both sides.7. Isolate x and solve.

ExampleExample

ExampleExample

Lesson 1 | Algebra


Page 12009Page 77

In this equation, a = 1. So we move to step two where we calculate ( b _ 2 )

2 .

b = –2, so ( b _ 2 ) 2 = ( –2 _ 2 )

2 = 1 .

Now we add this to each side of the equation and move the –24 to the right hand side of the equation:

x2 − 2x + 1 = 24 + 1 .

Now the square is completed and we can factorise:

\ x2 − 2x + 1 = 24 + 1

\ (x −1)2 = 25

\ x −1 = ±5

\ x = 1 ± 5

\ x = 6 or x = −4.

We isolate the constant term to get x2 − 2x = 24 . In this equation a = 1, so 4a = 4. So if we multiply each term by 4 we get:

4x2 − 8x = 96

Now if we add b2 = (−2)2 = 4 to both sides:

4x2 − 8x + 4 = 96 + 4 .

The square is completed and we can solve:

\ 4x2 − 8x + 4 = 96 + 4

\ (2x − 2)2 = 100

\ 2x − 2 = ±10

\ 2x = 2 ± 10

\ x = 1 ± 5

\ x = 6 or x = −4

You can choose which method is the easier for you to use. Always remember that working with an expression (y = ax2 + bx + c) is different to working with an equation (ax2 + bx + c = 0) .

Example 2

Solve for x if 2x2 − 3x − 1 = 0 by completing the square:

Method 1 Method 2Move the constant:

2x2 − 3x = 1

a = 2 so we divide by 2 throughout:

x2 − 3 _ 2 x = 1 _ 2

b = − 3 _ 2 , so ( b _ 2 ) 2 = (–

3 _ 2 _ 2 )

2

= (− 3 _ 4 ) 2 = 9 _ 16

We add this to both sides of the equation:

x2 − 3 _ 2 x + 9 _ 16 = 1 _ 2 + 9 _ 16

Now we solve for x:

x2 − 3 _ 2 x + 9 _ 16 = 1 _ 2 + 9 _ 16

\ (x − 3 _ 4 ) 2 = 8 + 9 _ 16 = 17 _ 16

\ x − 3 _ 4 = ± √ _ 17 _ 4

\ x = 3 ± √ _ 17 _ 4

Correct to two decimals:

\ x = 1,78 or x = −0,28

Move the constant:

2x2 − 3x = 1

Since a = 2 we see that 4a = 8. Since b = –3, we have b2 = (−3)2 = 9 .

So first multiply all terms by 4a, and then add the 9 to both sides:

(multiply by 8): 16x2 − 24x = 8

Add 9: 16x2 − 24x + 9 = 8 + 9

Now solve for x:

\ (4x − 3)2 = 17

\ 4x − 3 = ± √ _ 17

\ x = 3 ± √ _ 17 _ 4

Correct to two decimals:

\ x = 1,78 or x = −0,28

ExampleExample


Page 1Page 78

Example 3

Solve for x if 3x2 − 6x − 2 = 0 by completing the square:

Method 1 Method 2Isolate the constant term and divide by 3:

3x2 − 6x = 2

\ x2 − 2x = 2 _ 3

Complete the square and solve:

x2 − 2x = 2 _ 3

\ x2 − 2x + 1 = 2 _ 3 + 1 [ ( b _ 2 ) 2 = ( −2 _ 2 )

2 = 1]

\ ( x − 1)2 = 5 _ 3

\ x = 1 ± √ _ 5 _ 3 (surd form)

Isolate the constant term, and multiply by 4a.

Then add b2 to both sides:

3x2 − 6x = 2 (4a = 12; b2 = 36)

\ 36x2 − 72x + 36 = 24 + 36

\ (6x − 6)2 = 60

\ 6x − 6 = ± √ _ 60

\ x −1 = ± √ _ 60 _ 6

(

√ _ 60 _ 6 = √

_ 6 × √

_ 10 __ 6 =

√ _ 6 x

√ _ 10 __

√ _ 6 x

√ _ 6 = √

_ 10 _ 6 = √

_ 5 _ 3 )

\ x = 1 ± √ _ 5 _ 3

Activity 3

Solve for x in each of the following leaving your answers in simplest surd form and rounded to two decimal places where applicable: (use either of the methods for completing the square)

1. x2 − 6x + 3 = 0 2. x2 − 4 x + 5 = 0

3. 2x2 − x − 5 = 0 4. 3x2 + 4x = 7

ExampleExample

ActivityActivity

Lesson 1 | Algebra


Page 12009Page 79

5. 7 – 4x = 5x2

Solving quadratic inequalities

To solve an inequality in x means to find the set of values of x which satisfy the inequality. Solving an inequality is very similar to solving an equation but there are, however, two important differences:

(i) the algebraic rules for inequalities are different.

(ii) an equation usually has a finite number of ‘point’ solutions; an inequality is usually satisfied by whole intervals greater than or less than a specific value.

Linear inequalities

Let us look at three linear inequalities just to refresh our memories about what it is that we are working with:

Example 1

Solve for x: x + 4 ≥ 10

x + 4 ≥ 10

\ x ≥ 10 − 4

\ x ≥ 6

10x

y

1 2 3 4 5 6 7

8

1234567

–1–2–3–4 –1–2–3

–5

–4

8 9

1112131415

910

1112131415

For x ³ 6

y = 10

y = x + 4

In this area x + 4 ³ 10

Graphically: We sketch the line y = x + 4. Now we will search for the x values for which the y values on the line are bigger than or equal to 10. To do this, we also sketch the line y = 10. At the point of their intersection, we find the solution for this inequality. It happens at the point where x = 6. But because we are looking for any value of x that will make x + 4 bigger than or equal to 10, we see that x must be bigger than or equal to 6.

ExampleExample


Page 1Page 80

Example 2:

Solve for x in 3 ≤ x + 4 ≤ 8

We want to find a solution of the form a ≤ x ≤ b

So we want x in the middle, and not x + 4.

To obtain this we need to subtract 4 throughout.

3 ≤ x + 4 ≤ 8

\ 3 − 4 ≤ x + 4 − 4 ≤ 8 − 4

\ −1 ≤ x ≤ 4

So the solution set lies between negative one and four. Graphically the solution can be obtained as follows:

x

y

1 2 3 4 5 6 7

8

1234567

–1–2–3–4–1

–5 8

910

y = 8

y = x + 4

In this area 3 £ x + 4 £ 8

–1 £ x £ 4

y = 3

Here we need to consider three lines. They are y = x + 4, y = 3 and y = 8. Once we have sketched y = x + 4, we are searching where the y values on this line lie between 3 and 8. So we need to find the points of intersection where the lines y = 3 and y = 8 cross the line y = x + 4. We see that this happens at x = –1 and x = 4. Now since we want the set of values to lie between 3 and 8, we find our solution for x between –1 and 4.

Quadratic inequalities

In the examples to follow, the method for solving a quadratic inequality is vastly different from solving a linear inequality. Before we can solve any quadratic inequality, we must ensure that one side of the inequality is zero, by taking all terms to the LHS. After doing this, we ensure that all coefficients of x2 are positive. Finally, we have to factorise this new inequaliy fully. Our solutions will be based on the diagram alongside. We can see that they-values on the parabola are negative between its roots, and positive to either side of each root.

Example 1: Solve for x in x2 − 3x ≤ 4

We ensure that the one side of this inequality

is zero: x2 − 3x − 4 ≤ 0

Now we are ready to factorise:

\ (x − 4)(x + 1) ≤ 0

Now, in essence, this is a parabola (quadratic function) with two x-intercepts at x = −1 and at x = 4. So we can sketch the intercepts on a number line.

++ −

4–1

−1 ≤ x ≤ 4

ExampleExample

ExampleExample Since x2 − 3x − 4 ≤ 0 we are

interested in the less than

zero or ‘negative’ part of the

parabola.

Since x2 − 3x − 4 ≤ 0 we are

interested in the less than

zero or ‘negative’ part of the

parabola.

Lesson 1 | Algebra


Page 12009Page 81

Our answer means that y is negative below the x-axis. So for all the x-values between the two roots, y will be negative on the parabola. To the left of −1 and the right of 4, the y-values are positive. So our solution: –1 ≤ x ≤ 4.

Example 2: Solve for x in 3 + x − 2x2 ≤ 0

We make sure that the coefficient of x2 is positive:

2x2 − x − 3 ≥ 0 . Now we can factorise this inequality:

\ (2x − 3)(x + 1) ≥ 0

The roots here are x = −1 and x = 3 _ 2

++ −

32–1

\ x ≤ −1 or x ≥ 3 _ 2

Example 3: Solve for x in x2 + 2x + 6 ≥ 0

This inequality seems to have roots that are going to be difficult to find, since the factors of 6 will not add nor subtract to give the middle term 2. So we can complete the square to see what is happening here:

x2 + 2x + 6 = x2 + 2x + 1 + 5

= (x + 1)2 + 5

Now if we consider the fact that the square is now contained in a bracket, then we can argue that (x + 1)2 ≥ 0 for all x since anything that is squared is always greater than or equal to zero (positive). Now if we add 5 to each side, it remains positive.

\ (x + 1)2 + 5 ≥ 0 + 5

\ (x + 1)2 + 5 ≥ 5 for all x

So x2 + 2x + 6 = (x + 1)2 + 5 is always positive for all values of x.

\ x2 + 2x + 6 > 0 for all x Î ℝ

Graphically the situation is as follows:y

x0

6

(–1; 5)

Sometimes you can be required to show that a quadratic expression can only take on certain values, or possibly have a minimum or maximum value. Do not confuse these questions with solving an inequality. To find the extreme values,

ExampleExample

This inequality sign tells us we are looking for the ‘positive’ part of the parabola.This inequality sign tells us we are looking for the ‘positive’ part of the parabola.

ExampleExample


Page 1Page 82

you need to contain x in a bracket. We use completing of the square to do so. This will help us to point out that the squared part is always positive, and then interpret the rest of the expression. Let’s see how:

Example 4: Show that 2x2 − 4x + 5 > 0

We need to show that… not solve for x. So we want to work at containing the x in a squared bracket:

2x2 − 4x + 5 = 2 (x2 − 2x + 5 _ 2 )

= 2 (x2 − 2x + 1 + 3 _ 2 )

= 2 ((x2 − 2x + 1) + 3 _ 2 )

= 2 (( x + 1)2 + 3 _ 2 )

= 2(x + 1)2 + 3

Now: (x + 1)2 ≥ 0 for all x

So: 2(x + 1)2 ≥ 0 for all x

\ 2(x + 1)2 + 5 ≥ 5 for all x

\ 2(x + 1)2 + 5 > 0

This answer shows that the minimum value of the expression 2x2 − 4x + 5 is indeed 5.

Example 5: Show that x2 − 6x + 5 ≥ −4

Again we need not solve for x, but rather prove something about the expression. So we complete the square to get:

x2 − 6x + 5 = x2 − 6x + 9 − 4

= (x − 3)2 − 4

Now: (x − 3)2 ≥ 0 for all x

So: (x − 3)2 − 4 ≥ −4 for all x

Finally this indicates that the minimum value of x2 − 6x + 5 = (x − 3)2 − 4 is negative four.

Example 6: For which values of x will √ __ 3 + 2x − x2 be real.

We know that the square root of a negative number is non- real. So the number under the root must be positive for the root to be real.

\ 3 + 2x − x2 ≥ 0

\ x2 − 2x − 3 ≤ 0

\ (x − 3)(x + 1) ≤ 0

–1 3

+ +−

∴ –1 ≤ x≤ 3

ExampleExample

ExampleExample

ExampleExample

Lesson 1 | Algebra


Page 12009Page 83

Activity 4


1. x2 ≤ 9 2. 4x − 2x2 < 0

3. −7x2 + 6x + 1 < 0 4. 2x2 − x − 1 ≤ 0

5. x² + 2x + 1 ≥ 0 6. (x2 + 4)(x − 3) < 0

7. (x + 2)2(x2 + 4) > 0

8. Show that: − 1 _ 2 x2 + x + 4 ≤ 9 _ 2

9. Determine the minimum value of √ __

x2 − 8x + 25․

ActivityActivity


Page 1Page 84

Solving simultaneous equations

In this section we are required to solve two variables, so we must be supplied with two equations. This is a very important section in your work, as you will need to apply it later to graphs and also analytical geometry. We will always work with one equation that is linear, and one which is quadratic.

In our application, we must always substitute the linear equation into the quadratic equation. Avoid, if possible, forming fractions when you work with the linear equation.

When we solve simultaneous equations, we are finding the co-ordinates of the points where the graphs of the two curves cut one another.

METHOD:

1. Isolate one of the variables in the linear equation.

2. Substitute into the second equation.

3. Solve for the unknown.

4. Substitute back into the equation from step 1.

5. Solve the for the other unknown.

Example 1

Solve for x and y in:

x + 2y = 5 and 2y2 − xy − 4x2 = 8

Starting with the linear equation, we isolate one of the variables. Since x + 2y = 5, we will isolate the x since it has a coefficient of 1 and we will be avoiding forming unnecessary fractions.

So: x = 5 − 2y….(1) Now substitute into 2y2 − xy − 4x2 = 8 :

2y2 − (5 − 2y)y − 4(5 − 2y)2 = 8

\ 2y2 − 5y + 2y2 − 4(25 − 20y + 4y2) = 8

\ 4y2 − 5y −100 + 80y −16y2 − 8 = 0

\12y2 − 75y + 108 = 0 ( . _ . 3)

\ 4y2 − 25y + 36 = 0

\ (4y − 9)(y − 4) = 0

\ y = 9 _ 4 or y = 4

Now to solve for x since we have y, we substitute back into (1):

Then x = 5 − 2 ( 9 _ 4 ) or x = 5 − 2(4)

\ x = 5 − 9 _ 2 or x = 5 − 8

\ x = 10 − 9 _ 2 or x = −3

\ x = 1 _ 2

So x = 1 _ 2 ; y = 9 _ 4 or x = −3; y = 4

We can write the answers as:

(x; y) = ( 1 _ 2 ; 9 _ 4 ) or (x; y ) = (−3 ; 4)

•

•

ExampleExample

Lesson 1 | Algebra


Page 12009Page 85

Example 2

Solve for x and y in y = − 1 _ 2 x − 2 and y = x2 − 2x − 3

In both the equations, the y has already been isolated so we can just equate the two:

– 1 _ 2 x – 2 = x2 – 2x – 3

\ –x – 4 = 2x2 – 4x – 6

\ 2x2 – 3x – 2 = 0

\ (x – 2)(2x + 1) = 0

\ x = 2 or x = – 1 _ 2

If we now substitute these values back into the linear equation we get:

y = − 1 _ 2 (2) − 2 or y = − 1 _ 2 (− 1 _ 2 ) − 2

\ y = – 1 – 2 or y = 1 _ 4 – 2

\ y = –3 or y = –1 3 _ 4

So: (x; y) = (2; −3) or (x; y) = (− 1 _ 2 ; − 7 _ 4 )

Graphically the situation will be as follows:

2

(− ; −1 )

−

1 3 4 5

1

2

3

4

–1–2–3–4–5–1

–3

–4

–5

–6

x

y

y = x − 2x − 3

y = − x − 212

–212

34

− −

21 3

–3

–4

–5

3

(2 ; −3)

–6

Activity 5

Solve for x and y in each of the following:

1. x2 − 2xy − 5y2 = 3 = x − y

ExampleExample

ActivityActivity


Page 1Page 86

2. x + y = 9 and x2 + xy + y2 = 61

3. x − 2y = 3 and 4x2 − 5xy = 3 − 6y

4. (x − 1)2 + (y − 2)2 = 5 and 2x + y + 1 = 0

5. (y − 3)(x2 + 2) = 0

Working with equations that contain fractions

Here it is important to know that you may get rid of the denominator provided that you state the restrictions on the variable x. Do not keep the denominators when you solve the equation.

Lesson 1 | Algebra


Page 12009Page 87

Example 1

Solve for x in 2x _ 2x − 1 − 10 _ 3 = 1 – 2x _ 2x

Starting the restrictions on x in the denominators:

2x −1 ≠ 0 and 2x ≠ 0

\ x ≠ 1 _ 2 and x ≠ 0.

Solving for x:

3·2x(2x) − 10·2x(2x − 1) = 3·(2x − 1)(1 − 2x)

\ 12x2 − 40x2 + 20x = −3(2x −1)2

\ 12x2 − 40x2 + 20x = −12x2 + 12x − 3

\ −16x2 + 8x + 3 = 0

\ 16x2 − 8x − 3 = 0

\ (4x − 3)(4x + 1) = 0

\ x = 3 _ 4 or x = − 1 _ 4

Both these answers satisfy the restrictions and are therefore valid.

Example 2

Solve for x in

2x _ x − 1 + 3x _ 1 − x2 = 6 _ x + 1

We need to first factorise the denominator (1 − x2) completely. We can change the sign in the middle term so that

1 − x2 = − (x2 − 1) = − (x − 1)(x + 1) :

\ 2x _ x − 1 − 3x __ (x −1)(x + 1) = 6 _ x + 1

Restr: x ≠ ± 1 ; LCD (x −1)(x + 1)

\ 2x(x + 1) − 3x = 6(x −1)

\ 2x2 + 2x − 3x = 6x − 6

\ 2x2 − 7x + 6 = 0

\ (2x − 3)(x − 2) = 0

\ x = 3 _ 2 or x = 2

Example 3

Solve for x in

x2(x + 1) − 2x(x + 1) + (x + 1)

____ x2 − 1 = 0

Restr: x2 − 1 ≠ 0

\ x2 ≠ 1 ® x ≠ ±1

\ x2(x + 1) − 2x(x + 1) + (x + 1) = 0

\ (x + 1)(x2 − 2x + 1) = 0

\ (x + 1)(x − 1)2 = 0

\ x = −1 or x = 1

But x ≠ ± 1

Therefore there is no solution to this problem!

ExampleExample

Multiply through by the LCD : 3·2x(2x − 1) Multiply through by the LCD : 3·2x(2x − 1)

ExampleExample

ExampleExample

Factor out the common factor x + 1 Factor out the common factor x + 1


Page 1Page 88

Activity 6


1. 6 _ x + 1 − 1 _ x – 1 = 2 _ x 2. 4x _ x2 − 9

+ 3 _ x − 3 = 2 _ x + 3 + 1

3. 6 _ x + 1 – 3x _ 1 − x2 = 2x _ x − 1 4. x _ x + 1 – 2 _ 1 − x = x2 + 3 _

x2 – 1 + 9 _ 4

Solutions to Activities

Activity 1

1. x² − 7x + 10 = 0

\ (x − 2)(x − 5) = 0

\ x = 2 or x = 5

2. x² + 4x = 21

\ x² + 4x − 21 = 0

\ (x + 7)(x − 3) = 0

\ x = −7 or x = 3

3. 12x² − 6 = −17x −12

\ 12x2 + 17x + 6 = 0

\ (4x + 3)(3x + 2) = 0

\ x = − 3 _ 4 or x = − 2 _ 3

4. x² = 6 − 5x

\ x2 + 5x − 6 = 0

\(x + 6)(x − 1) = 0

\ x = −6 or x = 1

5. 6x2 + 11x + 3 = 0

\ (2x + 3)(3x + 1) = 0

\ x = – 3 _ 2 or x = – 1 _ 3

6. 8x2 – 14x + 3 = 0

\ (4x – 1)(2x – 3) = 0

\ x = 1 _ 4 or x = 3 _ 2

ActivityActivity

Lesson 1 | Algebra


Page 12009Page 89

7. 21x2 = 41x – 10

\ 21x2 – 41x + 10 = 0

\ (7x – 2)(3x – 5) = 0

\ x = 2 _ 7 or x = 5 _ 3

8. 3x2 + x = 14

\ 3x2 + x – 14 = 0

\ (3x + 7)(x – 2) = 0

\ x = – 7 _ 3 or x = 2

Activity 2

1. 2x − 5 = 7 _ x

\ 2x2 − 5x − 7 = 0

a = 2; b = −5; c = −7

\ x = 5 ± √ __

25 − 4(−14) ___ 4

\ x = 5 ± √ __ 25 + 56 __ 4

\ x = 5 ± √ _ 81 _ 4

\ x = 5 ± 9 _ 4

\ x = 7 _ 2 or x = −1

2. 1 _ 2 x2 − 3x + 0,7 = 0

\ 5x2 − 30x + 7 = 0

\ x = 30 ±

√ __

900 − 4(35) ___ 10

\ x = 30 ± √ _ 760 __ 10

\ x = 30 ± 2 √ _ 190 __ 10

x = 15 ± √ _ 190 __ 5

3. 4x2 − 20x + 9 = 0

a = 4; b = −20; c = 9

\ x = 20 ± √ __

400 − 4(36) ___ 8

\ x = 20 ± √ __ 400 −144 __ 8

\ x = 20 ± √ _ 256 __ 8

\ x = 20 ± 16 _ 8

\ x = 9 _ 2 or x = 1 _ 2

4. 2x2 + 4x + 3 = 0

a = 2 ; b = 4 ; c = 3

\ x = −b ± √ __ b2 − 4ac __ 2a

= −4 ± √ __ 16 − 4(6) __ 4

= −4 ± √ _ −8 __ 4

x is non-real


Page 1Page 90

Activity 3

1. x2 − 6x + 3 = 0

\ x2 − 6x + 9 = −3 + 9

\ (x − 3)2 = 6

\ x = 3 ± √ _ 6

\ x = 5,45 or x = 0,55

or

x2 − 6x + 3 = 0

\ 4x2 − 24x + 36 = −12 + 36

\ (2x − 6)2 = 24

\ 2x = 6 ± 2 √ _ 6

\ x = 3 ± √ _ 6

\ x = 5,45 or x = 0,55

2. x2 − 4x = −5

\ x2 − 4x + 4 = −5 + 4

\ (x − 2)2 = −1

\ x is non-real

or

x2 − 4x = −5

\ 4x2 −16x + 16 = −20 + 16

\ (2x − 4)2 = −4

\ x is non-real

3. 2x2 − x − 5 = 0

\ x2 − x _ 2 + 1 _ 16 = 5 _ 2 + 1 _ 16

\ (x − 1 _ 4 ) 2 = 41 _ 16

\ x = 1 ± √ _ 41 _ 4

\ x = 1,85 or x = −1,35

or

2x2 − x = 5

\ 16x2 − 8x + 1 = 40 + 1

\ (4x −1)2 = 41

\ x = 1 ± √ _ 41 _ 4

\ x = 1,85 or x = −1,35

4. 3x2 + 4x = 7

\ x2 + 4 _ 3 x + 16 _ 36 = 7 _ 3 + 16 _ 36

\ (x + 4 _ 6 ) 2 = 100 _ 36

\ x = − 4 ± 10 _ 6

\ x = − 7 _ 3 or x = 1

or

3x2 + 4x = 7

\ 36x2 + 48x + 16 = 84 + 16

\ (6x + 4)2 = 100

\ x = −4 ± 10 _ 6

\ x = − 7 _ 3 or x = 1

Lesson 1 | Algebra


Page 12009Page 91

5. 7 − 4x = 5x2

\ 5x2 + 4x = 7

\ x2 + 4 _ 5 x + 4 _ 25 = 7 _ 5 + 4 _ 25

\ (x + 2 _ 5 ) 2 = 39 _ 25

\ x = −2 ± √ _ 39 __ 5

\ x = 0,85 or x = −1,65

or

5x2 + 4x = 7

\ 100x2 + 80x + 16 = 140 + 16

\ (10x + 4)2 = 156

\ x = −4 ± 2 √ _ 39 __ 10

\ x = −2 ± √ _ 39 __ 5

\ x = 0,85 or x = −1,65

Activity 4

1. x2 ≤ 9

\ (x − 3)(x + 3) ≤ 0

+ +−3–3

\ −3 ≤ x ≤ 3

2. 4x − 2x2 < 0

\ 2x(x − 2) > 0

+ +−

20

\ x < 0 or x > 2

3. −7x2 + 6x + 1 < 0

\ 7x2 − 6x − 1 > 0

\ (7x + 1)(x − 1) > 0

+ +−1– 1_

7

\ x < − 1 _ 7 or x > 1

4. 2x2 − x − 1 ≤ 0

\ (2x + 1)(x − 1) ≤ 0


Page 1Page 92

+ +−

1– 1_2

\ − 1 _ 2 ≤ x ≤ 1

5. x² + 2x + 1 ≥ 0

\ (x + 1)2 ≥ 0

This is true for all x

6. (x2 + 4)(x − 3) < 0

Now : x2 ≥ 0 ® x2 + 4 ≥ 4

so for the result to be negative:

(x − 3) < 0

\ x < 3

7. (x + 2)2(x2 + 4) > 0

Now: x2 + 4 ≥ 4 > 0

So for the product to be positive:

(x + 2)2 > 0

\ x Î R − {−2}

8. − 1 _ 2 x2 + x + 4 = − 1 _ 2 (x2 − 2x − 8)

= − 1 _ 2 ((x2 − 2x + 1) − 9)

= − 1 _ 2 ( (x −1) 2 − 9)

= − 1 _ 2 (x −1)2 + 9 _ 2

So the maximum is 9 _ 2

9. √ __

x2 − 8x + 25 = √ ___

x2 − 8x + 16 + 9

= √ __ (x − 4)2 + 9

The minimum will happen when x = 4: i.e. √ _ 9 = 3.

Activity 5

1. x2 − 2xy − 5y2 = 3 … (1)

x = 3 + y … (2)

(1) ® (2) : (3 + y)2 − 2y(3 + y) − 5y2 = 3

\ y2 + 6y + 9 − 6y − 2y2 − 5y2 − 3 = 0

\ 6y2 − 6 = 0

\ y2 = 1

\ y = ±1

Now: x = 3 ± 1

\ x = 4 or x = 2

So (x; y) = (4; 1) or (x; y) = (2; −1)

Lesson 1 | Algebra


Page 12009Page 93

2. x = 9 − y ...(1) and x2 + xy + y2 = 61 …(2)

(1) ® (2) : (9 − y)2 + (9 − y)y + y2 = 61

\ y2 −18y + 81 + 9y − y2 + y2 − 61 = 0

\ y2 − 9y + 20 = 0

\(y − 5)(y − 4) = 0

\ y = 5 or y = 4

So: x = 9 − 5 or x = 9 − 4

\ x = 4 or x = 5

\(x; y) = (4; 5) or (x; y) = (5; 4)

3. x = 2y + 3 and 4x2 − 5xy = 3 − 6y

\4(2y + 3)2 − 5y(2y + 3) = 3 − 6y

\4(4y2 + 12y + 9) −10y2 −15y + 6y − 3 = 0

\16y2 + 48y + 36 −10y2 − 9y − 3 = 0

\6y2 + 39y + 33 = 0 ( . _ . 3)

\2y2 +13y + 11 = 0

\(2y +11)(y + 1) = 0

\ y = − 11 _ 2 or y = −1

So: x = 2 (− 11 _ 2 ) + 3 or x = 2(−1) + 3

\ x = −8 or x = 1

\(x; y) = (−8; − 11 _ 2 ) or (x; y) = (1; −1)

4. (x − 1)2 + (y − 2)2 = 5 and y = −2x −1

\ (x −1)2 + (−2x −1 − 2)2 = 5

\ x2 − 2x + 1 + 4x2 + 12x + 9 − 5 = 0

\5x2 + 10x + 5 = 0 ( . _ . 5)

\ x2 + 2x + 1 = 0

\(x + 1)2 = 0

\ x = −1

So: y = −2(−1) −1 = 1

\ (x; y) = (−1; 1)

5. (y − 3)(x2 + 2) = 0

\ y − 3 = 0 or x2 + 2 = 0

\ y = 3 only since x2 = −2 has no real solution.

Activity 6

1. 6 _ x + 1 − 1 _ x − 1 = 2 _ x

LCD: x(x − 1)(x + 1)

Restrictions: x ≠ 0 and x ≠ ±1

\ 6x(x − 1) − x(x + 1) = 2(x2 −1)


Page 1Page 94

\ 6x2 − 6x − x2 − x = 2x2 − 2

\ 3x2 − 7x + 2 = 0

\ (3x − 1)(x − 2) = 0

\ x = 1 _ 3 or x = 2

2. 4x __ (x − 3)(x + 3) + 3 _ x − 3 = 2 _ x + 3 + 1

LCD: (x − 3)(x + 3)

Restrictions: x ≠ ±3

\ 4x + 3(x + 3) = 2(x − 3) + (x − 3)(x + 3)

\ 4x + 3x + 9 = 2x − 6 + x2 − 9

\ x2 − 5x − 24 = 0

\ (x − 8)(x + 3) = 0

\ x = 8 or x = −3 n.a.

3. 6 _ x + 1 + 3x __ (x −1)(x + 1) = 2x _ x −1

LCD: (x − 1)(x + 1)

Restriction: x ≠ ±1

\ 6(x −1) + 3x = 2x(x + 1)

\ 6x − 6 + 3x = 2x2 + 2x

\ 2x2 − 7x + 6 = 0

\ (2x − 1)(x − 3) = 0

\ x = 1 _ 2 or x = 3

4. x _ x + 1 + 2 _ x − 1 = x2 + 3 __ (x −1)(x + 1) + 9 _ 4

LCD: 4(x − 1)(x + 1)

Restriction: x ≠ ±1

\ 4x(x − 1) + 2·4(x + 1) = 4(x2 + 3) + 9(x2 −1)

\ 4x2 − 4x + 8x + 8 = 4x2 + 12 + 9x2 − 9

\ 9x2 − 4x − 5 = 0

\ (9x + 5)(x −1) = 0

\ x = – 5 _ 9 or x = 1 n.a.

Note from earlier: − (1 − x2) = (x − 1)(x + 1) Note from earlier: − (1 − x2) = (x − 1)(x + 1)

− (1 − x) = (x − 1) − (1 − x) = (x − 1)

Documents

FACTORING AND SOLVING QUADRATIC LESSON EQUATIONS 11... · Solving quadratic equations by completing the square We can also solve equations of the form ax2 + bx + c = 0 by completing